\(\int x^2 (c+d x)^2 (a+b x^2)^p \, dx\) [13]

Optimal result

Integrand size = 20, antiderivative size = 144 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=-\frac {a c d \left (a+b x^2\right )^{1+p}}{b^2 (1+p)}+\frac {d^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {c d \left (a+b x^2\right )^{2+p}}{b^2 (2+p)}+\frac {1}{3} \left (c^2-\frac {3 a d^2}{5 b+2 b p}\right ) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right ) \] Output:

-a*c*d*(b*x^2+a)^(p+1)/b^2/(p+1)+d^2*x^3*(b*x^2+a)^(p+1)/b/(5+2*p)+c*d*(b* 
x^2+a)^(2+p)/b^2/(2+p)+1/3*(c^2-3*a*d^2/(2*b*p+5*b))*x^3*(b*x^2+a)^p*hyper 
geom([3/2, -p],[5/2],-b*x^2/a)/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.97 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {1}{15} \left (a+b x^2\right )^p \left (5 c^2 x^3 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {3 d \left (-\frac {5 c \left (a+b x^2\right ) \left (a-b (1+p) x^2\right )}{b^2}+d \left (2+3 p+p^2\right ) x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right )}{(1+p) (2+p)}\right ) \] Input:

Integrate[x^2*(c + d*x)^2*(a + b*x^2)^p,x]
 

Output:

((a + b*x^2)^p*((5*c^2*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])/ 
(1 + (b*x^2)/a)^p + (3*d*((-5*c*(a + b*x^2)*(a - b*(1 + p)*x^2))/b^2 + (d* 
(2 + 3*p + p^2)*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b 
*x^2)/a)^p))/((1 + p)*(2 + p))))/15
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {543, 27, 243, 53, 363, 279, 278, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 543

\(\displaystyle \int x^2 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+\int 2 c d x^3 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 27

\(\displaystyle \int x^2 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+2 c d \int x^3 \left (b x^2+a\right )^pdx\)

\(\Big \downarrow \) 243

\(\displaystyle \int x^2 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+c d \int x^2 \left (b x^2+a\right )^pdx^2\)

\(\Big \downarrow \) 53

\(\displaystyle \int x^2 \left (b x^2+a\right )^p \left (c^2+d^2 x^2\right )dx+c d \int \left (\frac {\left (b x^2+a\right )^{p+1}}{b}-\frac {a \left (b x^2+a\right )^p}{b}\right )dx^2\)

\(\Big \downarrow \) 363

\(\displaystyle \left (c^2-\frac {3 a d^2}{2 b p+5 b}\right ) \int x^2 \left (b x^2+a\right )^pdx+c d \int \left (\frac {\left (b x^2+a\right )^{p+1}}{b}-\frac {a \left (b x^2+a\right )^p}{b}\right )dx^2+\frac {d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 279

\(\displaystyle \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {3 a d^2}{2 b p+5 b}\right ) \int x^2 \left (\frac {b x^2}{a}+1\right )^pdx+c d \int \left (\frac {\left (b x^2+a\right )^{p+1}}{b}-\frac {a \left (b x^2+a\right )^p}{b}\right )dx^2+\frac {d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 278

\(\displaystyle c d \int \left (\frac {\left (b x^2+a\right )^{p+1}}{b}-\frac {a \left (b x^2+a\right )^p}{b}\right )dx^2+\frac {1}{3} x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {3 a d^2}{2 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 2009

\(\displaystyle c d \left (\frac {\left (a+b x^2\right )^{p+2}}{b^2 (p+2)}-\frac {a \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}\right )+\frac {1}{3} x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c^2-\frac {3 a d^2}{2 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {d^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

Input:

Int[x^2*(c + d*x)^2*(a + b*x^2)^p,x]
 

Output:

(d^2*x^3*(a + b*x^2)^(1 + p))/(b*(5 + 2*p)) + c*d*(-((a*(a + b*x^2)^(1 + p 
))/(b^2*(1 + p))) + (a + b*x^2)^(2 + p)/(b^2*(2 + p))) + ((c^2 - (3*a*d^2) 
/(5*b + 2*b*p))*x^3*(a + b*x^2)^p*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2 
)/a)])/(3*(1 + (b*x^2)/a)^p)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int x^{2} \left (d x +c \right )^{2} \left (b \,x^{2}+a \right )^{p}d x\]

Input:

int(x^2*(d*x+c)^2*(b*x^2+a)^p,x)
 

Output:

int(x^2*(d*x+c)^2*(b*x^2+a)^p,x)
 

Fricas [F]

\[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2} \,d x } \] Input:

integrate(x^2*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="fricas")
 

Output:

integral((d^2*x^4 + 2*c*d*x^3 + c^2*x^2)*(b*x^2 + a)^p, x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (121) = 242\).

Time = 8.84 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.78 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {a^{p} c^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {a^{p} d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + 2 c d \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate(x**2*(d*x+c)**2*(b*x**2+a)**p,x)
 

Output:

a**p*c**2*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**p 
*d**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + 2*c*d*Pi 
ecewise((a**p*x**4/4, Eq(b, 0)), (a*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3 
*x**2) + a*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2* 
b**3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x**2* 
log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(x - sqrt 
(-a/b))/(2*b**2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, -1)) 
, (-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x**2*(a 
 + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x**2) 
**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b**2* 
p**2 + 6*b**2*p + 4*b**2), True))
 

Maxima [F]

\[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2} \,d x } \] Input:

integrate(x^2*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^2*(b*x^2 + a)^p*x^2, x)
 

Giac [F]

\[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2} \,d x } \] Input:

integrate(x^2*(d*x+c)^2*(b*x^2+a)^p,x, algorithm="giac")
 

Output:

integrate((d*x + c)^2*(b*x^2 + a)^p*x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\int x^2\,{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^2 \,d x \] Input:

int(x^2*(a + b*x^2)^p*(c + d*x)^2,x)
 

Output:

int(x^2*(a + b*x^2)^p*(c + d*x)^2, x)
 

Reduce [F]

\[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^p \, dx=\text {too large to display} \] Input:

int(x^2*(d*x+c)^2*(b*x^2+a)^p,x)
 

Output:

( - 8*(a + b*x**2)**p*a**2*c*d*p**3 - 36*(a + b*x**2)**p*a**2*c*d*p**2 - 4 
6*(a + b*x**2)**p*a**2*c*d*p - 15*(a + b*x**2)**p*a**2*c*d - 6*(a + b*x**2 
)**p*a**2*d**2*p**3*x - 18*(a + b*x**2)**p*a**2*d**2*p**2*x - 12*(a + b*x* 
*2)**p*a**2*d**2*p*x + 4*(a + b*x**2)**p*a*b*c**2*p**4*x + 22*(a + b*x**2) 
**p*a*b*c**2*p**3*x + 38*(a + b*x**2)**p*a*b*c**2*p**2*x + 20*(a + b*x**2) 
**p*a*b*c**2*p*x + 8*(a + b*x**2)**p*a*b*c*d*p**4*x**2 + 36*(a + b*x**2)** 
p*a*b*c*d*p**3*x**2 + 46*(a + b*x**2)**p*a*b*c*d*p**2*x**2 + 15*(a + b*x** 
2)**p*a*b*c*d*p*x**2 + 4*(a + b*x**2)**p*a*b*d**2*p**4*x**3 + 14*(a + b*x* 
*2)**p*a*b*d**2*p**3*x**3 + 14*(a + b*x**2)**p*a*b*d**2*p**2*x**3 + 4*(a + 
 b*x**2)**p*a*b*d**2*p*x**3 + 4*(a + b*x**2)**p*b**2*c**2*p**4*x**3 + 24*( 
a + b*x**2)**p*b**2*c**2*p**3*x**3 + 49*(a + b*x**2)**p*b**2*c**2*p**2*x** 
3 + 39*(a + b*x**2)**p*b**2*c**2*p*x**3 + 10*(a + b*x**2)**p*b**2*c**2*x** 
3 + 8*(a + b*x**2)**p*b**2*c*d*p**4*x**4 + 44*(a + b*x**2)**p*b**2*c*d*p** 
3*x**4 + 82*(a + b*x**2)**p*b**2*c*d*p**2*x**4 + 61*(a + b*x**2)**p*b**2*c 
*d*p*x**4 + 15*(a + b*x**2)**p*b**2*c*d*x**4 + 4*(a + b*x**2)**p*b**2*d**2 
*p**4*x**5 + 20*(a + b*x**2)**p*b**2*d**2*p**3*x**5 + 35*(a + b*x**2)**p*b 
**2*d**2*p**2*x**5 + 25*(a + b*x**2)**p*b**2*d**2*p*x**5 + 6*(a + b*x**2)* 
*p*b**2*d**2*x**5 + 48*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p 
+ 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3 
*d**2*p**6 + 360*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 1...